1

Crystal Structures

1.1 Crystal Structure and Symmetry Groups

Although everyone has an intuitive idea of what a solid is, we will consider
(in this book) only materials with a well-dened crystal structure. What we
mean by a well-dened crystal structure is an arrangement of atoms in a lattice
such that the atomic arrangement looks absolutely identical when viewed from
two dierent points that are separated by a lattice translation vector. A few
denitions are useful.
Lattice
A lattice is an innite array of points obtained from three primitive translation vectors a1 , a2 , a3 . Any point on the lattice is given by
n = n1 a1 + n2 a2 + n3 a3 .

(1.1)

Translation Vector
Any pair of lattice points can be connected by a vector of the form
Tn1 n2 n3 = n1 a1 + n2 a2 + n3 a3 .

(1.2)

The set of translation vectors form a group called the translation group of the
lattice.
Group
A set of elements of any kind with a set of operations, by which any two
elements may be combined into a third, satisfying the following requirements
is called a group:

The product (under group multiplication) of two elements of the group

belongs to the group.

1 Crystal Structures

Fig. 1.1. Translation operations in a two-dimensional lattice

The associative law holds for group multiplication.

The identity element belongs to the group.
Every element in the group has an inverse which belongs to the group.

Translation Group
The set of translations through any translation vector Tn1 n2 n3 forms a group.
Group multiplication consists in simply performing the translation operations
consecutively. For example, as is shown in Fig. 1.1, we have T13 = T03 +
T10 . For the simple translation group the operations commute, i.e., Tij Tkl =
Tkl Tij for, every pair of translation vectors. This property makes the group
an Abelian group.
Point Group
There are other symmetry operations which leave the lattice unchanged. These
are rotations, reections, and the inversion operations. These operations form
the point group of the lattice. As an example, consider the two-dimensional
square lattice (Fig. 1.2). The following operations (performed about any lattice
point) leave the lattice unchanged.

E: identity
R1 , R3 : rotations by 90
R2 : rotation by 180
mx , my : reections about x-axis and y-axis, respectively
m+ , m : reections about the lines x = y

The multiplication table for this point group is given in Table 1.1. The operations in the rst column are the rst (right) operations, such as m+ in
R1 m+ = my , and the operations listed in the rst row are the second (left)
operations, such as R1 in R1 m+ = my .

1.1 Crystal Structure and Symmetry Groups

Fig. 1.2. The two-dimensional square lattice

Table 1.1. Multiplication table for the group 4 mm. The rst (right) operations,
such as m+ in R1 m+ = my , are listed in the rst column, and the second (left)
operations, such as R1 in R1 m+ = my , are listed in the rst row
Operation
1

E =E
R11 = R3
R21 = R2
R31 = R1
m1
x = mx
m1
y = my
m1
+ = m+
m1
= m

R1

R2

R3

mx

my

m+ m

E
R3
R2
R1
mx
my
m+
m

R1
E
R3
R2
m+
m
my
mx

R2
R1
E
R3
my
mx
m
m+

R3
R2
R1
E
m
m+
mx
my

mx
m+
my
m
E
R2
R3
R1

my
m
mx
m+
R2
E
R1
R3

m+
my
m
mx
R1
R3
E
R2

m
mx
m+
my
R3
R1
R2
E

Fig. 1.3. Identity operation on a two-dimensional square

The multiplication table can be obtained as follows:

Label the corners of the square (Fig. 1.3).

Operating with a symmetry operation simply reorders the labeling. For
example, see Fig. 1.4 for symmetry operations of m+ , R1 , and mx .

Therefore, R1 m+ = my . One can do exactly the same for all other products, for example, such as my R1 = m+ . It is also very useful to note what happens to a point (x, y) under the operations of the point group (see Table 1.2).
Note that under every group operation x x or y and y y or x.

1 Crystal Structures

Fig. 1.4. Point symmetry operations on a two-dimensional square

Table 1.2. Point group operations on a point (x, y)

Operation

R1

R2

R3

mx

my

m+

x
y

x
y

y
x

x
y

y
x

x
y

x
y

y
x

y
x

Fig. 1.5. The two-dimensional rectangular lattice

The point group of the two-dimensional square lattice is called 4 mm. The
notation denotes the fact that it contains a fourfold axis of rotation and two
mirror planes (mx and my ); the m+ and m planes are required by the existence of the other operations. Another simple example is the symmerty group
of a two-dimensional rectangular lattice (Fig. 1.5). The symmetry operations
are E, R2 , mx , my , and the multiplication table is easily obtained from that
of 4 mm. This point group is called 2 mm, and it is a subgroup of 4 mm.
Allowed Rotations
Because of the requirement of translational invariance under operations of the
translation group, the allowed rotations of the point group are restricted to
certain angles. Consider a rotation through an angle about an axis through
some lattice point (Fig. 1.6). If A and B are lattice points separated by a
primitive translation a1 , then A (and B ) must be a lattice point obtained
by a rotation through angle about B (or about A). Since A and B

1.1 Crystal Structure and Symmetry Groups

Fig. 1.6. Allowed rotations of angle about an axis passing through some lattice
points A and B consistent with translational symmetry
Table 1.3. Allowed rotations of the point group
p

cos

1
0

1
1
2

0
1
2

n (= |2/|)

0 or 2
2

6
2

4
2

3
2

1
6
4
3
2

are lattice points, the vector B A must be a translation vector. Therefore, we

have
|B A | = pa1 ,
(1.3)



 
where p is an integer. But |B A | = a1 + 2a1 sin 2 = a1 2a1 cos .
Solving for cos gives
1p
cos =
.
(1.4)
2
Because 1 cos 1, we see that p must have only the integral values -1,
0, 1, 2, 3. This gives for the possible values of listed in Table 1.3.
Although only rotations of 60 , 90 , 120, 180 , and 360 are consistent
with translational symmetry, rotations through other angles are obtained
in quasicrystals (e.g., vefold rotations). The subject of quasicrystals, which
do not have translational symmetry under the operations of the translation
group, is an interesting modern topic in solid state physics which we will not
discuss in this book.
Bravais Lattice
If there is only one atom associated with each lattice point, the lattice is called
Bravais lattice. If there is more than one atom associated with each lattice
point, the lattice is called a lattice with a basis. One atom can be considered
to be located at the lattice point. For a lattice with a basis it is necessary to
give the locations (or basis vectors) for the additional atoms associated with
the lattice point.

1 Crystal Structures

a
a

cos =

a
2a

Fig. 1.7. Construction of the WignerSeitz cell of a two-dimensional centered

rectangular lattice. Note that cos = a1 /2a2

Primitive Unit Cell

From the three primitive translation vectors a1 , a2 , a3 , one can form a parallelepiped that can be used as a primitive unit cell. By stacking primitive unit
cells together (like building blocks) one can ll all of space.
WignerSeitz Unit Cell
From the lattice point (0, 0, 0) draw translation vectors to neighboring lattice points (to nearest, next nearest, etc., neighbors). Then, draw the planes
which are perpendicular bisectors of these translation vectors (see, for example, Fig. 1.7). The interior of these intersecting planes (i.e., the space closer
to (0, 0, 0) than to any other lattice point) is called the WignerSeitz unit
cell.
Space Group
For a Bravais lattice, the space group is simply the product of the operations
of the translation group and of the point group. For a lattice with a basis,
there can be other symmetry operations. Examples are glide planes and screw
axes; illustration of each is shown in Figs. 1.8 and 1.9, respectively.
Glide Plane
In Fig. 1.8, each unit cell contains six atoms and T1/2 mx is a symmertry
operation even though neither T1/2 nor mx is an operation of the symmetry
group.

1.1 Crystal Structure and Symmetry Groups

Fig. 1.8. Glide plane of a two-dimensional lattice. Each unit cell contains six atoms

Fig. 1.9. Screw axis. Unit cell contains three layers and T1 is the smallest
translation. Occupied sites are shown by solid dots

Screw Axis
In Fig. 1.9, T1/3 R120 is a symmetry operation even though T1/3 and R120
themselves are not.
Two-Dimensional Bravais Lattices
There are only ve dierent types of two-dimensional Bravais lattices.
1. Square lattice: primitive (P) one only
It has a = b and = = 90 .
2. Rectangular: primitive (P) and centered (C) ones
They have a = b but = = 90 .

10

1 Crystal Structures

3. Hexagonal: primitive (P) one only

It has a = b and = 120( = 60 ).
4. Oblique: primitive (P) one only
It has a = b and = 90 .
Three-Dimensional Bravais Lattices
There are 14 dierent types of three-dimensional Bravais lattices.
1. Cubic: primitive (P), body centered (I), and face centered (F) ones.
For all of these a = b = c and = = = 90 .
2. Tetragonal: primitive and body centered (I) ones.
For these a = b = c and = = = 90 . One can think of them as cubic
lattices that have been stretched (or compressed) along one of the cube
axes.
3. Orthorombic: primitive (P), body centered (I), face centered (F), and base
centered (C) ones.
For all of these a = b = c but = = = 90 . These can be thought
of as cubic lattices that have been stretched (or compressed) by dierent
amounts along two of the cube axes.
4. Monoclinic: primitive (P) and base centered (C) ones.
For these a = b = c and = = 90 = . These can be thought of as
orthorhombic lattices which have suered shear distortion in one direction.
5. Triclinic: primitive (P) one.
This has the lowest symmetry with a = b = c and = = .
6. Trigonal:
It has a = b = c and = = = 90 < 120 . The primitive cell is a
rhombohedron. The trigonal lattice can be thought of as a cubic lattice
which has suered shear distortion.
7. Hexagonal: primitive (P) one only.
It has a = b = c and = = 90 , = 120.

1.2 Common Crystal Structures

1. Cubic
(a) Simple cubic (sc): Fig. 1.10
For simple cubic crystal the lattice constant is a and the volume per
atom is a3 . The nearest neighbor distance is also a, and each atom has
six nearest neighbors. The primitive translation vectors are a1 = a
x,
a2 = a
y, a3 = a
z.
(b) Body centered cubic (bcc): Fig. 1.11
If we take a unit cell as a cube of edge a, there are two atoms
per cell (one at (0, 0, 0) and one at 12 , 12 , 12 ). The atomic volume

is 12 a3 , and the nearest neighbor distance is 23 a. Each atom has

1.2 Common Crystal Structures

11

Fig. 1.10. Crystallographic unit cell of a simple cubic crystal of lattice constant a

Fig. 1.11. Crystallographic unit cell of a body centered cubic crystal of lattice
constant a

eight nearest neighbors. The primitive translations can be taken as

a1 = 12 a (
x + y + z), a2 = 12 a (
x + y + z), and a3 = 12 a (
x y + z).
The parallelepiped formed by a1 , a2 , a3 is the primitive unit cell (containing a single atom), and there is only one atom per primitive unit
cell.
(c) Face centered cubic (fcc): Fig. 1.12
If we take a unit cell as a cube of edge a, there are four atoms per cell;
1
1
8 of one at each of the eight corners and 2 of one on each of the six
a3
faces. The volume per atom is 4 ; the nearest neighbor distance is a2 ,
and each atom has 12 nearest neighbors. The primitive unit cell is the
parallelepiped formed from the primitive translations a1 = 12 a (
x + y),
a2 = 12 a (
y + z), and a3 = 12 a (
z+x
).
All three cubic lattices have the cubic group as their point group.
Because the primitive translations are dierent, the simple cubic, bcc,
and fcc lattices have dierent translation groups.

12

1 Crystal Structures

Fig. 1.12. Crystallographic unit cell of a face centered cubic crystal of lattice
constant a

Fig. 1.13. Crystallographic unit cell of a simple hexagonal crystal of lattice

constants a1 , a2 , and c

2. Hexagonal
(a) Simple hexagonal: See Fig. 1.13.
(b) Hexagonal close packed (hcp):
This is a non-Bravais lattice. It contains two atoms per primitive unit
cell of thesimple hexagonal lattice, one at (0,0,0) and the second
at 13 , 23 , 12 . The hexagonal close packed crystal can be formed by
stacking the rst layer (A) in a hexagonal array as is shown in Fig. 1.14.
Then, the second layer (B) is formed by stacking atoms in the alternate
triangular holes on top of the rst layer. This gives
another hexagonal

layer displaced from the rst layer by 13 , 23 , 12 . Then the third layer
is placed above the rst layer (i.e., at (0,0,1)). The stacking is then
repeated ABABAB . . .. If one stacks ABCABC . . ., where C is the
hexagonal array obtained by stacking the third layer in the other set
of triangular holes above the set B (instead of the set A), one gets
an fcc lattice. The closest possible packing of the hcp atoms occurs

1.2 Common Crystal Structures

13

Fig. 1.14. Stacking of layers A and B in a hexagonal close packed crystal of lattice
constants a1 , a2 , and c

3.

4.

5.

6.

7.

8.


when ac = 8/3 1.633. We leave this as an exercise for the reader.
Zn crystalizes in a hcp lattice with a = 2.66
A and c = 4.96
A giving
c
c

1.85,
larger
than
the
ideal
value.
a
a
Zincblende Structure This is a non-Bravais lattice.It is an FCC with two
atoms per primitive unit cell located at (0,0,0) and 14 , 14 , 14 . The structure
can be viewed as two interpenetrating fcc lattices displaced by one fourth
of the body diagonal. Examples of the zincblende structure are ZnS (cubic
phase), ZnO (cubic phase), CuF, CuCl, ZnSe, CdS, GaN (cubic phase),
InAs, and InSb. The metallic ions are on one sublattice, the other ions on
the second sublattice.
Diamond Structure This structure is identical to the zincblende structure,
except that there are two identical atoms in the unit cell. This
 struc
ture (unlike zincblende) has inversion symmetry about the point 18 , 18 , 18 .
Diamond, Si, Ge, and gray tin are examples of the diamond structure.
Wurtzite Structure This structure is a simple hexagonal
with

 lattice


four atoms per unit cell, located at (0,0,0), 13 , 23 , 12 , 0, 0, 38 , and 13 , 23 , 78 .
It can be pictured
 as consisting of two interpenetrating hcp lattices separated by 0, 0, 38 . In the wurtzite phase of ZnS, the Zn atoms sit on one hcp
lattice and the S atoms on the other. ZnS, BeO, ZnO (hexagonal phase),
CdS, GaN (hexagonal phase), and AlN are materials that can occur in the
wurtzite structure.
Sodium Chloride Structure It consists of a face centered cubic lattice with
a 1basis
of two unlike atoms per primitive unit cell, located at (0,0,0) and
1 1
2 , 2 , 2 . In addition to NaCl, other alkali halide salts like LiH, KBr, RbI
form crystals with this structure.
Cesium Chloride Structure It consists of a simple cubic lattice with two
atoms per unit cell, located at (0,0,0) and 12 , 12 , 12 . Besides CsCl, CuZn
(-brass), AgMg, and LiHg occur with this structure.
Calcium Fluoride Structure It consists of a face centered cubic lattice with
three atoms
 per primitive

 unit cell. The Ca ion is located at (0,0,0), the F
atoms at 14 , 14 , 14 and 34 , 34 , 34 .

14

1 Crystal Structures

Fig. 1.15. Stacking of layers A and B in a graphite structure

9. Graphite Structure This structure consists of a simple hexagonal

 2 1 lat
tice
with
four
atoms
per
primitive
unit
cell,
located
at
(0,0,0),
,
,
0
,
3
3

1 2 1

1
0, 0, 2 , and 3 , 3 ,2 . It can
 be pictured as two interpenetrating HCP lattices separated by 0, 0, 12 . It, therefore, consists of tightly bonded planes
(as shown in Fig. 1.15) stacked in the sequence ABABAB . . .. The individual planes are very tightly bound, but the interplanar binding is rather
weak. This gives graphite its well-known properties, like easily cleaving
perpendicular to the c-axis.
Miller Indices
Miller indices are a set of three integers that specify the orientation of a crystal
plane. The procedure for obtaining Miller indices of a plane is as follows:
1. Find the intercepts of the plane with the crystal axes.
2. Take the reciprocals of the three numbers.
3. Reduce (by multiplying by the same number) this set of numbers to the
smallest possible set of integers.
As an example, consider the plane that intersects the cubic axes at A1 , A2 , A3
as shown
in Fig. 1.16.
Then xi ai = OAi . The reciprocals of (x1 , x2 , x3 )


1
1
are x1
,
and
the Miller indices of the plane are (h1 h2 h3 ) =
,
x
,
x
 1 1 1 2 1 3
p
 x1 , x2 , x3 , where (h1 h2 h3 ) are the smallest possible set of integers
p
p
p
x1 , x2 , x3

Indices of a Direction
A direction in the lattice can be specied by a vector V = u1 a1 + u2 a2 + u3 a3 ,
or by the set of integers [u1 u2 u3 ] chosen to have no common integral factor. For
cubic lattices the plane (h1 h2 h3 ) is perpendicular to the direction [h1 h2 h3 ],
but this is not true for general lattices.
Packing Fraction
The packing fraction of a crystal structure is dened as the ratio of the volume
of atomic spheres in the unit cell to the volume of the unit cell.

1.3 Reciprocal Lattice

15

Fig. 1.16. Intercepts of a plane with the crystal axes

Examples
1. Simple cubic lattice:
We take the atomic radius as R = a2 (then neighboring atoms just touch).
The packing fraction p will be given by

p=

4
3

 a 3
2
a3

0.52
6

2. Body centered cubic lattice:

 
Here, we take R = 12 23 a , i.e., half the nearest neighbor distance. For
the non-primitive cubic cell of edge a, we have two atoms per cell giving

p=

2 43

a3

3
a 3
4

3 0.68
8

1.3 Reciprocal Lattice

If a1 , a2 , a3 are the primitive translations of some lattice, we can dene the
vectors b1 , b2 , b3 by the condition
ai bj = 2ij ,

(1.5)

where ij = 0 if i is not equal to j and ii = 1. It is easy to see that

bi = 2

aj ak
,
ai (aj ak )

(1.6)

16

1 Crystal Structures

where i, j, and k are dierent. The denominator ai (aj ak ) is simply the

volume v0 of the primitive unit cell. The lattice formed by the primitive translation vectors b1 , b2 , b3 is called the reciprocal lattice (reciprocal to the lattice
formed by a1 , a2 , a3 ), and a reciprocal lattice vector is given by
Gh1 h2 h3 = h1 b1 + h2 b2 + h3 b3 .

(1.7)

Useful Properties of the Reciprocal Lattice

1. If r = n1 a1 + n2 a2 + n3 a3 is a lattice vector, then we can write r as

(r bi ) ai .
(1.8)
r=
i

2. The lattice reciprocal to b1 , b2 , b3 is a1 , a2 , a3 .

3. A vector Gh from the origin to a point (h1 , h2 , h3 ) of the reciprocal lattice
is perpendicular to the plane with Miller indices (h1 h2 h3 ).
4. The distance from the origin to the rst lattice plane (h1 h2 h3 ) is
1
d (h1 h2 h3 ) = 2 |Gh | . This is also the distance between neighboring
{h1 h2 h3 } planes.
The proof of 3 is established by demonstrating that Gh is perpendicular to
the plane A1 A2 A3 shown in Fig. 1.16. This must be true if Gh is perpendic
ular to both A1 A2 and to A2 A3 . But A1 A2 = OA2 OA1 = p ha22 ha11 .
Therefore,


a2
a1
Gh A1 A2 = (h1 b1 + h2 b2 + h3 b3 ) p
,
(1.9)

h2
h1
which vanishes. The same can be done for A2 A3 . The proof of 4 is established
by noting that
a1 Gh
d(h1 h2 h3 ) =
.

h1 |Gh |
The rst factor is just the vector OA1 for the situation where p = 1, and
the second factor is a unit vector perpendicular to the plane (h1 h2 h3 ). Since
1
a1 Gh = 2h1 , it is apparent that d(h1 h2 h3 ) = 2 |Gh | .

1.4 Diraction of X-Rays

Crystal structures are usually determined experimentally by studying how
the crystal diracts waves. Because the interatomic spacings in most crystals
are of the order of a few
As (1
A = 108 cm), the maximum information
can most readily be obtained by using waves whose wave lengths are of that

1.4 Diraction of X-Rays

17

order of magnitude. Electromagnetic, electron, or neutron waves can be used

to study diraction by a crystal. For electromagnetic waves, E = h, where
E is the energy of the photon, = c is its frequency and its wave length,
and h is Plancks constant. For = 108 cm, c = 3 1010 cm/s and h =
6.6 1027 erg s, the photon energy is equal to roughly 2 108 ergs or
1.24 104 eV. Photons of energies of tens of kilovolts are in the X-ray range.
For electron waves, p = h  6.6 1019 g cm/s when = 108 cm. This
2

p
, where me  0.9 1027 g, of 2.4 1010 ergs or roughly
gives E = 2m
e
150 eV. For neutron waves, we need simply replace me by mn = 1.67 1024 g
to obtain E = 1.3 1013 ergs  0.08 eV. Thus neutron energies are of the
order of a tenth of an eV. Neutron scattering has the advantages that the low
energy makes inelastic scattering studies more accurate and that the magnetic
moment of the neutron allows the researcher to obtain information about
the magnetic structure. It has the disadvantage that high intensity neutron
sources are not as easily obtained as X-ray sources.

1.4.1 Bragg Reection

We have already seen that we can discuss crystal planes in a lattice structure.
Assume that an incident X-ray is specularly reected by a set of crystal planes
as shown in Fig. 1.17. Constructive interference occurs when the dierence in
path length is an integral number of wave length . It is clear that this occurs
when
2d sin = n,

(1.10)

where d is the interplane spacing, is the angle between the incident beam
and the crystal planes, as is shown on the gure, and n is an integer. Equation
(1.10) is known as Braggs law.
1.4.2 Laue Equations
A slightly more elegant discussion of diraction from a crystal can be obtained
as follows:
INCIDENT
WAVE

REFLECTED
WAVE

Fig. 1.17. Specular reection of X-rays by a set of crystal planes separated by a

distance d

18

1 Crystal Structures

Fig. 1.18. Scattering of X-rays by a pair of atoms in a crystal

1. Let s0 be a unit vector in the direction of the incident wave, and s be a

unit vector in the direction of the scattered wave.
2. Let R1 and R2 be the position vectors of a pair of atoms in a Bravais
lattice, and let r12 = R1 R2 .
Let us consider the waves scattered by R1 and by R2 and traveling dierent
path lengths as shown in Fig. 1.18. The dierence in path length is | R2 A
BR1 |. But this is clearly equal to |r12 s r12 s0 |. We dene S as S = s s0 ;
then the dierence in path length for the two rays is given by
= |r12 S| .

(1.11)

For constructive interference, this must be equal to an integral number of

wave length. Thus, we obtain
r12 S = m,

(1.12)

where m is an integer and is the wave length. To obtain constructive interference from every atom in the Bravais lattice, this must be true for every
lattice vector Rn . Constructive interference will occur only if
Rn S = integer

(1.13)

for every lattice vector Rn in the crystal. Of course there will be dierent
integers for dierent Rn in general. Recall that
Rn = n1 a1 + n2 a2 + n3 a3 .

(1.14)

The condition (1.13) is obviously satised if

ai S = phi ,

(1.15)

where hi is the smallest set of integers and p is a common multiplier. We can

obviously express S as
S = (S a1 ) b1 + (S a2 ) b2 + (S a3 ) b3 .

(1.16)

1.4 Diraction of X-Rays

19

Fig. 1.19. Relation between the scattering vector S = s s0 and the Bragg angle

Therefore, condition (1.13) is satised and constructive interference from every

lattice site occurs if
S = p (h1 b1 + h2 b2 + h3 b3 ) ,

(1.17)

S
= pGh ,

(1.18)

or

where Gh is a vector of the reciprocal lattice. Equation (1.18) is called the

Laue equation.
Connection of Laue Equations and Braggs Law
From (1.18) S must be perpendicular to the planes with Miller indices
(h1 h2 h3 ). The distance between two planes of this set is
d(h1 h2 h3 ) =

2
=p .
|Gh |
|S|

(1.19)

We know that S is normal to the reection plane PP with Miller indices
(h1 h2 h3 ). From Fig. 1.19, it is apparent that |S| = 2 sin . Therefore, (1.19)
can be written by
2d(h1 h2 h3 ) sin = p,
where p is an integer. According to Laues equation, associated with any
reciprocal lattice vector Gh = h1 b1 + h2 b2 + h3 b3 , there is an X-ray reection
satisfying the equation 1 S = pGh , where p is an integer.
1.4.3 Ewald Construction
This is a geometric construction that illustrates how the Laue equation works.
The construction goes as follows: See Fig. 1.20.
1. From the origin O of the reciprocal lattice draw the vector AO of length
1 parallel to s0 and terminating on O.
2. Construct a sphere of radius 1 centered at A.

20

1 Crystal Structures

Fig. 1.20. Ewald construction for diraction peaks

If this sphere intersects a point B of the reciprocal lattice, then AB = s is

in a direction in which a diraction maximum occurs. Since A1 O = s01 and
s0
A1 B1 = s1 , S1 = s
1 = OB1 is a reciprocal lattice vector and satises the
Laue equation. If a higher frequency X-ray is used, 2 , A2 , and B2 replace 1 ,
A1 , and B1 . For a continuous spectrum with 1 2 , all reciprocal lattice
1
points between the two sphere (of radii 1
1 and 2 ) satisfy Laue equation
for some frequency in the incident beam.
Wave Vector
It is often convenient to use the set of vectors Kh = 2Gh . Then, the Ewald
construction gives
q0 + Kh = q,
(1.20)
0 and q = 2
are the wave vectors of the incident and
where q0 = 2
s
s
scattered waves. Equation (1.20) says that wave vector is conserved up to 2
times a vector of the reciprocal lattice.
1.4.4 Atomic Scattering Factor
It is the electrons of an atom that scatter the X-rays since the nucleus is so
heavy that it hardly moves in response to the rapidly varying electric eld of
the X-ray. So far, we have treated all of the electrons as if they were localized
at the lattice point. In fact, the electrons are distributed about the nucleus of
the atom (at position r = 0, the lattice point) with a density (r). If you know

1.4 Diraction of X-Rays

21

Fig. 1.21. Path dierence between waves scattered at O and those at r

the wave function (r1 , r2 , . . . , rz ) describing the z electrons of the atom, (r)
is given by
 z








(r) =
(r ri ) = (r1 , . . . , rz ) 
(r ri ) (r1 , . . . , rz ) .


i=1
i=1
(1.21)
Now, consider the dierence in path length between waves scattered at O
and those scattered at r (Fig. 1.21).
= r (
s s0 ) = r S.

(1.22)

The phase dierence is simply 2

times , the dierence in path length.
Therefore, the scattering amplitude will be reduced from the value obtained
by assuming all the electrons were localized at the origin O by a factor z 1 f ,
where f is given by

2i
f = d3 r (r) e rS .
(1.23)
This factor is called the atomic scattering factor. If (r) is spherically symmetric we have
1
2i
f=
2r2 dr d(cos )(r)e Sr cos .
(1.24)
0

Recall that S = 2 sin , where is the angle between s0 and the reecting
plane PP of Fig. 1.19. Dene as 4
sin ; then f can be expressed as

sin r
.
(1.25)
f=
dr4r2 (r)
r
0
If is much larger than the atomic radius, r is much smaller than unity
wherever (r) is nite. In that case sinrr  1 and f z, the number of
electrons.

22

1 Crystal Structures

1.4.5 Geometric Structure Factor

So far we have considered only a Bravais lattice. For a non-Bravais lattice the
scattered amplitude depends on the locations and atomic scattering factors
of all the atoms in the unit cell. Suppose a crystal structure contains atoms
at positions rj with atomic scattering factors fj . It is not dicult to see that
this changes the scattered amplitude by a factor

2i
F (h1 , h2 , h3 ) =
fj e rj S(h1 h2 h3 )
(1.26)
j

for the scattering from a plane with Miller indices (h1 h2 h3 ). In (1.26) the
position vector rj of the jth atom can be expressed in terms of the primitive
translation vectors ai

rj =
ji ai .
(1.27)
i

For example, in a hcp lattice r1 = (0, 0, 0) and r2 = ( 13 , 23 , 12 ) when expressed

in
terms of the primitive translation vectors. Of course, S(h1 h2 h3 ) equal to
i hi bi , where bi are primitive translation vectors in the reciprocal lattice.
Therefore, 2i
rj S(h1 h2 h3 ) is equal to 2i (j1 h1 + j2 h2 + j3 h3 ), and the
structure amplitude F (h1 , h2 , h3 ) can be expressed as


fj e2i i ji hi .
(1.28)
F (h1 , h2 , h3 ) =
j

If all of the atoms in the unit cell are identical (as in diamond, Si, Ge, etc.)
all of the atomic scattering factors fj are equal, and we can write
F (h1 , h2 , h3 ) = f S(h1 h2 h3 ).

(1.29)

The S(h1 h2 h3 ) is called the geometric structure amplitude. It depends only

on crystal structure, not on the atomic constituents, so it is the same for all
hcp lattices or for all diamond lattices, etc.
Example
A useful demonstration of the geometric structure factor can be obtained by
considering a bcc lattice as a simple cubic lattice with two atoms in the simple
cubic unit cell located at (0,0,0) and ( 12 , 12 , 12 ). Then
S(h1 h2 h3 ) = 1 + e2i( 2 h1 + 2 h2 + 2 h3 ) .
1

(1.30)

If h1 + h2 + h3 is odd, e
= 1 and S(h1 h2 h3 ) vanishes. If h1 + h2 +
h3 is even, S(h1 h2 h3 ) = 2. The reason for this eect is that the additional
planes (associated with the body centered atoms) exactly cancel the scattering
amplitude from the planes made up of corner atoms when h1 + h2 + h3 is odd,
but they add constructively when h1 + h2 + h3 is even.
The scattering amplitude depends on other factors (e.g. thermal motion
and zero point vibrations of the atoms), which we have neglected by assuming
a perfect and stationary lattice.
i(h1 +h2 +h3 )

1.4 Diraction of X-Rays

23

1.4.6 Experimental Techniques

We know that constructive interference from a set of lattice planes separated
by a distance d will occur when
2d sin = n,

(1.31)

where is the angle between the incident beam and the planes that are scattering, is the X-ray wave length, and n is an integer. For a given crystal
the possible values of d are xed by the atomic spacing, and to satisfy (1.31),
one must vary either or over a range of values. Dierent experimental
methods satisfy (1.31) in dierent ways. The common techniques are (1) the
Laue method, (2) the rotating crystal method, and (3) the powder method.
Laue Method
In this method a single crystal is held stationary in a beam of continuous wave
length X-ray radiation (Fig. 1.22). Various crystal planes select the appropriate wave length for constructive interference, and a geometric arrangement of
bright spots is obtained on a lm.
Rotating Crystal Method
In this method a monochromatic beam of X-ray is incident on a rotating
single crystal sample. Diraction maxima occur when the sample orientation
relative to the incident beam satises Braggs law (Fig. 1.23).
Powder Method
Here, a monochromatic beam is incident on a nely powdered specimen. The
small crystallites are randomly oriented with respect to the incident beam,
so that the reciprocal lattice structure used in the Ewald construction must
be rotated about the origin of reciprocal space through all possible angles.
This gives a series of spheres in reciprocal space of radii K1 , K2 , . . . (we
include the factor 2 in these reciprocal lattice vectors) equal to the smallest,

SPOT
PATTERN

X - RAY
BEAM

SAMPLE
COLLIMATOR
FILM

Fig. 1.22. Experimental arrangement of the Laue method

24

1 Crystal Structures

MONOCHROMATIC
X-RAY BEAM

FILM
ROTATABLE
SAMPLE

Fig. 1.23. Experimental arrangement of the rotating crystal method

X - RAY BEAM

DIFFRACTION
RINGS
POWDER
SAMPLE
FILM

Fig. 1.24. Experimental arrangement of the powder method

next smallest, etc. reciprocal lattice vectors. The sequence of values

sin(i /2)
sin(1 /2)

Ki
give the ratios of K
for the crystal structure. This sequence is determined
1
by the crystal structure. Knowledge of the X-ray wave length = 2
k allows
determination of the lattice spacing (Fig. 1.24).

1.5 Classication of Solids

1.5.1 Crystal Binding
Before considering even in a qualitative way how atoms bind together to
form crystals, it is worthwhile to review briey the periodic table and the
ground state congurations of atoms. The single particle states of electrons
moving in an eective central potential (which includes the attraction of the
nucleus and some average repulsion associated with all other electrons) can
be characterized by four quantum numbers: n, the principal quantum number
takes on the values 1, 2, 3, . . ., l, the angular momentum quantum number
takes on values 0, 1, . . . , n 1; m, the azimuthal quantum number (projection
of l onto a given direction) is an integer satisfying l m l; and , the
spin quantum number takes on the values 21 .

1.5 Classication of Solids

25

The energy of the single particle orbital is very insensitive to m and (in
the absence of an applied magnetic eld), but it depends strongly on n and l.
Of course, due to the Pauli principle only one electron can occupy an orbital
with given n, l, m, and . The periodic table is constructed by making an
array of slots, with l value increasing from l = 0 as one moves to the left, and
the value of n + l increasing as one moves down. (Table 1.4) Of course, the
correct number of slots must be allowed to account for the spin and azimuthal
degeneracy 2(2l + 1) of a given l value. One then begins lling the slots from
the top left, moving to the right, and then down when all slots of a given
(n + l) value have been used up. See Table 1.4, which lists the atoms (H,
He, . . .) and their atomic numbers in the appropriate slots. As the reader can
readily observe, H has one electron, and it will occupy the n = 1, l = 0(1s)
state. Boron has ve electrons and they will ll the (1s) and 2s states with the
fth electron in the 2p state. Everything is very regular until Cr and Cu. These
two elements have ground states in which one 4s electron falls into the 3d shell,
giving for Cr the atomic conguration (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)1 (3d)5 ,
and for Cu the atomic conguration (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)1 (3d)10 .
Other exceptions occur in the second transition series (the lling of the 4d
levels) and in the third transition series (lling the 5d levels), and in the rare
earth series (lling the 4f and 5f levels). Knowing this table allows one to
write down the ground state electronic conguration of any atom. Note that
the inert gases He, Ne, Kr, Rn, complete the shells n = 1, n = 2, n = 3, and
n = 4, respectively. Ar and Xe are inert also; they complete the n = 3 shell
(except for 3d electrons), and n = 4 shell (except for 4f electrons), respectively. Na, K, Rb, Cs, and Fr have one weakly bound s electron outside these
closed shell congurations; Fl, Cl, Br, I and At are missing one p electron from
the closed shell congurations. The alkali metals easily give up their loosely
bound s electrons, and the halogens readily attract one p electron to give a
closed shell conguration. The resulting Na+ Cl ions form an ionic bond
which is quite strong. Atoms like C, Si, Ge, and Sn have an (np)2 (n + 1 s)2
conguration. These four valence electrons can be readily shared with other
atoms in covalent bonds, which are also quite strong.1 Compounds like GaAs,

In Table 1.4, we note exceptions (i)(viii):

i Dropping a 4s electron into the 3d shell while lling 3d shell (Cr, Cu)
ii Dropping a 5s electron into the 4d shell while lling 4d shell (Nb, Mo, Ru, Rh,
Ag)
iii Dropping both 5s electrons into the 4d shell while lling 4d shell (Pd)
iv Dropping both 6s electrons into the 5d shell while lling 5d shell (Pt)
v Dropping one 6s electron into the 5d shell while lling 5d shell (Au)
vi Adding one 5d electron before lling the entire 4f shell (La, Gd)
vii Adding one 6d electron before lling the entire 5f shell (Ac, Pa, U, Cm, Cf)
viii Adding two 5d electrons before lling the entire 5f shell (Th, Bk)

65
Tb

66
Dy

67 68 69 70 71 72
Ho Er Tm Yb Lu Hf

i
ii
iii
iv
v
vi
vii
viii

73
Ta

74
W

75
Re

76
Os

77
Ir

Dropping a 4s electron into the 3d shell while lling 3d shell (Cr, Cu)
Dropping a 5s electron into the 4d shell while lling 4d shell (Nb, Mo, Ru, Rh, Ag)
Dropping both 5s electrons into the 4d shell while lling 4d shell (Pd)
Dropping both 6s electrons into the 5d shell while lling 5d shell (Pt)
Dropping one 6s electron into the 5d shell while lling 5d shell (Au)
Adding one 5d electron before lling the entire 4f shell (La, Gd)
Adding one 6d electron before lling the entire 5f shell (Ac, Pa, U, Cm, Cf)
Adding two 5d electrons before lling the entire 5f shell (Th, Bk)

We note exceptions (i)(viii):

89(vii) 90(viii) 91(vii) 92(vii) 93 94 95 96(vii) 97(viii) 98(vii) 99 100 101 102
Np Pu Am Cm
Es F Md No
Ac
Th
Pa
U
Bk
Cf

61 62 63 64(vi)
Pm Sm Eu
Gd

87
Fr

60
Nd

78(iv) 79(v) 80 81 82 83 84 85 86
Pt
Au Hg Tl Pb Bi Po At Rn

59
Pr

57(vi)
La

58
Ce

55
Cs

39 40 41(ii) 42(ii) 43 44(ii) 45(ii) 46(iii) 47(ii) 48 49 50 51 52 53 54

Y Zr Nb
Rh
Pd
Mo Tc Ru
Ag Cd In Sn Sb Te I Xe

28
Ni

l = 3

27
Co

37
Rb

26
Fe

29(i) 30 31 32 33 34 35 36
Cu Zn Ga Ge As Se Br Kr

21 22
Sc Ti

24(i) 25
Cr Mn

19
K

13 14 15 16 17 18
Al Si P S Cl Ar
23
V

11
Na

6 7 8 9 10
C N O F Ne

5
B

l =2

4
Be

3
Li

l = 1

88
Ra

56
Ba

38
Sr

20
Ca

12
Mg

2
He

1
H

l = 0

8
(5f, 6d, 7p, 8s)

7
(4f, 5d, 6p, 7s)

6
(4d, 5p, 6s)

5
(4p, 5s)

4
(3p, 4s)

3
(2p, 3s)

2
(2s)

1
(1s)

n+l

Table 1.4. Ground state electron congurations in a periodic table: Note that n = 1, 2, 3, . . .; l < n; m = l, l + 1, . . . , l; = 12

26
1 Crystal Structures

1.6 Binding Energy of Ionic Crystals

27

GaP, GaSb, or InP, InAs, InSb, etc., are formed from column III and column
V constituents. With the partial transfer of an electron from As to Ga, one
obtains the covalent bonding structure of Si or Ge as well as some degree of
ionicity as in NaCl. Metallic elements like Na and K are relatively weakly
bound. Their outermost s electrons become almost free in the solid and act as
a glue holding the positively charged ions together. The weakest bonding in
solids is associated with weak Van der Waals coupling between the constituent
atoms or molecules. To give some idea of the binding energy of solids, we will
consider the binding of ionic crystals like NaCl or CsCl.

1.6 Binding Energy of Ionic Crystals

The binding energy of ionic crystals results primarily from the electrostatic
interaction between the constituent ions. A rough order of magnitude estimate
of the binding energy per molecule can be obtained by simply evaluating

2
4.8 1010 esu
e2
V
=
 8 1012 ergs 5eV.
=
R0
2.8 108 cm
A, the
Here, R0 is the observed interatomic spacing (which we take as 2.8
spacing in NaCl). The experimentally measured value of the binding energy
of NaCl is almost 8 eV per molecule, so our rough estimate is not too bad.
Interatomic Potential
For an ionic crystal, the potential energy of a pair of atoms i, j can be taken
to be
e2

ij =
+ n.
(1.32)
rij
rij
Here, rij is the distance between atoms i and j. The sign depends on
whether the atoms are like (+) or unlike (). The rst term is simply the
Coulomb potential for a pair of point charges separated by rij . The second
term accounts for core repulsion. The atoms or ions are not point charges, and
when a pair of them gets close enough together their core electrons can repel
one another. This core repulsion is expected to decrease rapidly with increasing rij . The parameters and n are phenomenological; they are determined
from experiment.
Total Energy
The total potential energy is given by
U=

1
ij .
2
i=j

(1.33)

28

1 Crystal Structures

It is convenient to dene i , the potential energy of the ith atom as

i =




ij .

(1.34)

Here, the prime on the sum implies that the term i = j is omitted. It is
apparent from symmetry considerations that i is independent of i for an
innite lattice, so we can drop the subscript i. The total energy is then
U=

1
2N = N ,
2

(1.35)

where 2N is the number of atoms and N is the number of molecules.

It is convenient in evaluating to introduce a dimensionless parameter pij
dened by pij = R1 rij , where R is the distance between nearest neighbors.
In terms of pij , the expression for is given by
=



 n e2 
p

(pij )1 .
ij
Rn j
R j

(1.36)

Here, the primes on the summations denote omission of the term i = j. We

dene the quantities


An =
pn
(1.37)
ij ,
j

and
=

(pij )

(1.38)

The and An are properties of the crystal structure; is called the Madelung
constant. The internal energy of the crystal is given by N , where N is the
number of molecules. The internal energy is given by


An
e2
U =N n
.
(1.39)
R
R
 
At the equilibrium separation R0 , U
R R0 must vanish. This gives the result

e2
An
=
.
n
R0
nR0

Therefore, the equilibrium value of the internal energy is


1
e2
U0 = N 0 = N
.
1
R0
n

(1.40)

(1.41)

1.6 Binding Energy of Ionic Crystals

29

Compressibility
The best value of the parameter n can be determined from experimental data
on the compressibility . is dened by the negative of the change in volume
per unit change in pressure at constant temperature divided by the volume.


1 V
=
.
(1.42)
V P T
The subscript T means holding temperature T constant, so that (1.42) is the
isothermal compressibility. We will show that at zero temperature
 2
U
1
=V
.
(1.43)
V 2 T =0
Equation (1.43) comes from the thermodynamic relations
F = U T S,

(1.44)

dU = T dS P dV.

(1.45)

and
By taking the dierential of (1.44) and making use of (1.45), one can see that
dF = P dV SdT.
From (1.46) we have

F
V

P =


=V

2F
V 2

(1.47)

Equation (1.42) can be written as


 2
P
F
1
=V
.
= V
V T
V 2 T
But at T = 0, F = U so that

(1.46)

(1.48)

(1.49)
T =0

is the inverse of the isothermal compressibility at T = 0. We can write the

R
1

= V
volume V as 2N R3 and use V
R = 6N R2 R in (1.39) and (1.43). This
gives
1
T =0 =

e2
(n 1),
18R04

or

(1.50)

18R04
.
(1.51)
e2
From the experimental data on NaCl, the best value for n turns out to be
9.4.
n=1+

30

1 Crystal Structures

Evaluation of the Madelung Constant

For simplicity let us start with a linear chain. Each positive (+) atom has two
neighbors, which are negative () atoms, at p01 = 1. Therefore,
=




p1
ij



1 1 1
= 2 1 + + ... .
2 3 4

(1.52)

If you remember that the power series expansion for ln(1 + x) is given by
n

2
3
4
n=1 (x)
= x x2 + x3 x4 + and is convergent for x 1, it is
n
apparent that
= 2 ln 2.
(1.53)
If we attempt the same approach for NaCl, we obtain
=

6
8
6
12
+ + .
1
2
3 2

(1.54)

This is taking six opposite charge nearest neighbors at a separation

of one
nearest neighbor distance, 12 same charge next nearest neighbors at 2 times
that distance, etc. It is clear that the series in (1.54) converges very poorly.
The convergence can be greatly improved by using a dierent counting procedure in which one works with groups of ions which form a more or less
neutral array. The motivation is that the potential of a neutral assembly of
charges falls o much more quickly with distance than that of a charged
assembly.
Evjens Method
We will illustrate Evjens method 2 by considering a simple square lattice in
two dimensions with two atoms per unit cell, one at (0, 0) and one at ( 12 , 12 ).
The crystal structure is illustrated in Fig. 1.25. The calculation is carried out
as follows:
1. One considers the charges associated with dierent shells where the rst
shell is everything inside the rst square, the second is everything outside
the rst but inside the second square, etc.
2. An ion on a face is considered to be half inside and half outside the square
dened by that face; a corner atom is one quarter inside and three quarters
outside.
3. The total Madelung constant is given by = 1 + 2 + 3 + , where
j is the contribution from the ith shell.
As an example, let us evaluate the total charge on the rst few shells. The
rst shell has four atoms on faces, all with the opposite charge to the atom
2

H.M. Evjen, Phys. Rev. 39, 675 (1932).

1.6 Binding Energy of Ionic Crystals

31

PRIMITIVE
UNIT CELL

Fig. 1.25. Evjen method for a simple square lattice in two dimensions

at the origin and four corner atoms all with the same charge as the atom at
the origin. Therefore, the charge of shell number one is


1
1
Q1 = 4
4
= 1.
(1.55)
2
4
Doing the same for the second shell gives





3
1
1
1
1
Q2 = 4
4
4
+8
4
= 0.
2
4
2
2
4

(1.56)

Here the rst two terms come from the remainder of the atoms on the outside
of the rst square; the next three terms come from the atoms on the inside of
the second square. To get 1 and 2 we simply divide the individual charges
by their separations from the origin. This gives
1 =

2 =

4 ( 12 ) 4 ( 14 )
 1.293,
1
2

4 ( 12 ) 4 ( 34 ) 4 ( 12 ) 8 ( 12 ) 4 ( 14 )

+ +  0.314.
1
2
2
5
2 2

(1.57)

(1.58)

This gives  1 + 2 1.607. The readers should be able to evaluate 3

for themselves.
Madelung Constant for Three-Dimensional Lattices
For a three-dimensional crystal, Evjens method is essentially the same with
the exception that

32

1 Crystal Structures

Fig. 1.26. Central atom and the rst cube of the Evjen method for the NaCl
structure

1. The squares are replaced by cubes.

2. Atoms on the face of a cube are considered to be half inside and half outside
the cube; atoms on the edge are 14 inside and 34 outside, and corner atoms
are 18 inside and 78 outside.
We illustrate the case of the NaCl structure as an example in the three
dimensions (see Fig. 1.26).
For 1 we obtain
1 =

6 ( 12 ) 12 ( 14 ) 8 ( 18 )

+  1.456.
1
2
3

(1.59)

For 2 we have the following contributions:

1. Remainder of the contributions from the atoms on the rst cube
6 (1)
12 ( 3 )
8 (7)
= 12 24 + 38
2. Atoms on the interior of faces of the second cube
6 (1)
6(4) ( 1 )
6(4) ( 1 )
= 22 + 5 2 6 2
3. Atoms on the interior of edges of the second cube
12 ( 1 )

12(2) ( 1 )

= 84 + 9 4
4. Atoms on the interior of the coners of the second cube
8 ( 18 )
= 12
Adding them together gives




12
1
7
3
12
3
6
9
 0.295.
2 = 3 +
+ +
+ +
2
2
3
5
6
8 3
12
(1.60)

1.6 Binding Energy of Ionic Crystals

33

Thus, to the approximation  1 + 2 we nd that  1.752. The exact

result for NaCl is = 1.747558 . . ., so Evjens method gives a surprisingly
accurate result after only two shells.
Results of rather detailed evaluations of for several dierent crystal structures are (NaCl) = 1.74756, (CsCl) = 1.76267, (zincblende) = 1.63806,
(wurtzite) = 1.64132. The NaCl structure occurs much more frequently than
the CsCl structure. This may seem a bit surprising since (CsCl) is about 1%
larger than (NaCl). However, core repulsion accounts for about 10% of the
binding energy (see (1.41)). In the CsCl structure, each atom has eight nearest
neighbors instead of the six in NaCl. This should increase the core repulsion
by something of the order of 25% in CsCl. Thus, we expect about 2.5% larger
contribution (from core repulsion) to the binding energy of CsCl. This negative
contribution to the binding energy more than compensates the 1% dierence
in Madelung constants.

34

1 Crystal Structures

Problems
1.1. Demonstrate that
(a) The reciprocal lattice of a simple cubic lattice is simple cubic.
(b) The reciprocal lattice of a body centered cubic lattice is a face centered
cubic lattice.
(c) The reciprocal lattice of a hexagonal lattice is hexagonal.
1.2. Determine the packing fraction of
(a) A simple cubic lattice
(b) A face centered lattice
(c) A body centered lattice
(d) The diamond structure
(e) A hexagonal close packed lattice with an ideal
(f) The graphite structure with an ideal ac .

c
a

ratio

1.3. The Bravais lattice of the diamond structure is fcc with two carbon atoms
per primitive unit cell. If one of the two basis atoms is at (0,0,0), then the
other is at ( 14 , 14 , 14 ).
(a) Illustrate that a reection through the (100) plane followed by a nonprimitive translation through [ 14 , 14 , 14 ] is a glide-plane operation for the
diamond structure.
(b) Illustrate that a fourfold rotation about an axis in diamond parallel to
the x-axis passing through the point (1, 14 , 0) (the screw axis) followed
by the translation [ 14 , 0, 0] parallel to the screw axis is a screw operation
for the diamond structure.
1.4. Determine the group multiplication table of the point group of an
equilateral triangle.
1.5. CsCl can be thought of as a simple cubic lattice with two dierent atoms
[at (0, 0, 0) and ( 12 , 12 , 12 )] in the cubic unit cell. Let f+ and f be the atomic
scattering factors of the two constituents.
(a) What is the structure amplitude F (h1 , h2 , h3 ) for this crystal?
(b) An X-ray source has a continuous spectrum with wave
k
 numbers

satisfying: k is parallel to the [110] direction and 21/2 2

|k|

a
 
3 21/2 2
,
where
a
is
the
edge
distance
of
the
simple
cube.
Use the
a
Ewald construction for a plane that contains the direction of incidence
to show which reciprocal lattice points display diraction maxima.
(c) If f+ = f , which of these maxima disappear?
1.6. A simple cubic structure is constructed in which two planes of A atoms
followed by two planes of B atoms alternate in the [100] direction.

1.6 Binding Energy of Ionic Crystals

35

1. What is the crystal structure (viewed as a non-Bravais lattice with four

atoms per unit cell)?
2. What are the primitive translation vectors of the reciprocal lattice?
3. Determine the structure amplitude F (h1 , h2 , h3 ) for this non-Bravais lattice.
1.7. Powder patterns of three cubic crystals are found to have their rst four
diraction rings at the values given below:
A
B

30
21
30

35
29
50

50
36
60

60
42
74

The crystals are monatomic, and the observer believes that one is body
centered, one face centered, and one is a diamond structure.
1. What structures are the crystals A, B, and C?
2. The wave length of the incident X-ray is 0.95
A. What is the length of
the cube edge for the cubic unit cell in A, B, and C, respectively?
1.8. Determine the ground state atomic congurations of C(6), O(8), Al(13),
Si(16), Zn(30), Ga(31), and Sb(51).
1.9. Consider 2N ions in a linear chain with alternating e charges and a
repulsive potential ARn between nearest neighbors.
1. Show that, at the equilibrium separation R0 , the internal energy becomes


1
N e2
.
1
U (R0 ) = 2 ln 2
R0
n
2. Let the crystal be compressed so that R0 R0 (1 ). Show that the work
done per unit length in compressing the crystal can be written 12 C 2 , and
determine the expression for C.
1.10. For a BCC and for an FCC lattice, determine the separations between
nearest neighbors, next nearest neighbors, . . . down to the 5th nearest neighbors. Also determine the separations between nth nearest neighbors (n =
1, 2, 3, 4, 5) in units of the cube edge a of the simple cube.

36

1 Crystal Structures

Summary
In this chapter rst we have introduced basic geometrical concepts useful in
describing periodic arrays of objects and crystal structures both in real and
reciprocal spaces assuming that the atoms sit at lattice sites.
A lattice is an innite array of points obtained from three primitive
translation vectors a1 , a2 , a3 . Any point on the lattice is given by
n = n1 a1 + n2 a2 + n3 a3 .
Any pair of lattice points can be connected by a vector of the form
Tn1 n2 n3 = n1 a1 + n2 a2 + n3 a3 .
Well dened crystal structure is an arrangement of atoms in a lattice such
that the atomic arrangement looks absolutely identical when viewed from two
dierent points that are separated by a lattice translation vector. Allowed
types of Bravais lattices are discussed in terms of symmetry operations
both in two and three dimensions. Because of the requirement of translational invariance under operations of the lattice translation, the rotations of
60 , 90 , 120 , 180 , and 360 are allowed.
If there is only one atom associated with each lattice point, the lattice
of the crystal structure is called Bravais lattice. If more than one atom is
associated with each lattice point, the lattice is called a lattice with a basis.
If a1 , a2 , a3 are the primitive translations of some lattice, one can dene a set
of primitive translation vectors b1 , b2 , b3 by the condition
ai bj = 2ij ,
where ij = 0 if i is not equal to j and ii = 1. It is easy to see that
bi = 2

aj ak
,
ai (aj ak )

where i, j, and k are dierent. The lattice formed by the primitive translation vectors b1 , b2 , b3 is called the reciprocal lattice (reciprocal to the lattice
formed by a1 , a2 , a3 ), and a reciprocal lattice vector is given by
Gh1 h2 h3 = h1 b1 + h2 b2 + h3 b3 .
Simple crystal structures and principles of commonly used experimental
methods of wave diraction are also reviewed briey. Connection of Laue
equations and Braggs law is shown. Classication of crystalline solids are
then discussed according to conguration of valence electrons of the elements
forming the solid.